Sockets - Shirley #22
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jadevance
reviewed
Apr 24, 2019
| # Returns a new array to that contains elements in the intersection of the two input arrays | ||
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n), because the lookup for a hash is constant, and the times we do a lookup depends on the length of the longer array |
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I believe this is an implementation of insertion sort, which is quadratic in time complexity.
Source 1: Comparison Sort
Source 2: Time Complexity
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n), because the lookup for a hash is constant, and the times we do a lookup depends on the length of the longer array | ||
| # Space complexity: O(n). where n is the length of the shortest array, since all elements of the shortest array are plaeced in the hash |
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If it is quadratic, space complexity is O(1).
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| intersection_array = [] | ||
| longer_array.each do |element| | ||
| intersection_array.push(element) if shorter_array_hash.include?(element) |
| shorter_array_hash = {} | ||
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| until index > shorter_array.length | ||
| shorter_array_hash[shorter_array[index]] = 1 |
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I am not sure if you need a hash here. If the intersection of the arrays had multiple values that were in common (example:
test1 = [1,2,3,4,5,5,5]
test2 = [5,5,5,5,5]
shorter_array_hash would be reset to a value of 1 each time for every iteration of 5. I think an array would work for your purposes.
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